Which of the following is a true for subtropical jet streams

Not sure what to do can you help

NeTakaya

sorry i don't know the answer i'm really sorry

7 0

2 years ago

What is the mass of KOH found in a 785 mL of a 1.43 M solution of KOH

My name is Ann [436]

Molarity is defined as the ratio of number of moles to the volume of solution in litres.

The mathematical expression is given as:

$Molarity = \frac{Number of moles}{volume of solution in litres}$

Here, molarity is equal to 1.43 M and volume is equal to 785 mL.

Convert mL into L

As, 1 mL = 0.001 L

Thus, volume = $785\times 0.001$ = 0.785 L

Rearrange the formula of molarity in terms of number of moles:

$Number of moles =molarity\times volume of solution in litres$

n = $1.43 M \times 0.785 L$

= 1.12255 mole

Now, Number of moles = $\frac{mass in g}{molar mass}$

Molar mass of potassium hydroxide = 56.10 g/mol

1.12255 mole = $\frac{mass in g}{56.10 g/mol}$

mass in g = $1.12255 mole\times 56.10 g/mol$

= 62.97 g

Hence, mass of $KOH$ = 62.97 g

7 0

2 years ago

If 17.4 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentra

Verizon [17]

NH3 +HCl ----> NH4Cl

moles of HCl used = (0.8 x 17.4) /1000= 0.0139 moles
by use of reacting ratio between HCl to NH4Cl which is 1:1 therefore the moles of NH4Cl is also = 0.0139 moles
molar concentration = moles /volume in liters

molar concentration is therefore= (0.0139/5) x1000 = 2.7 M

8 0

2 years ago

Read 2 more answers

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.